Question:

For $f(x)$, observe that $(a^x - a^{-x})^2 + (a^x + a^{-x})^2 = 2(a^{2x} + a^{-2x})$ for any base $a$. This means the first two columns of $f(x)$ satisfy a linear dependence relation. Specifically, $C_1 + C_2 = 2(a^{2x} + a^{-2x})$ for each row. The determinant $f(x)$ vanishes because the three columns become linearly dependent—each row has the form where $C_1 + C_2$ is constant while $C_3$ is always 1. Therefore $f(x) = 0$ for all $x$, making $f(4) = 0$. For $g(x)$, performing row operations: $R_2 - R_1$ and $R_3 - R_1$ yields a determinant that factors as $(x-1)h(x)$ where $h(x)$ is a quadratic. Since $f(x) = 0$ identically, the equation $f(x) = g(x)$ becomes $g(x) = 0$, which has three solutions (one being $x=1$ with multiplicity considerations from the quadratic factor).